2024 CGMO

Proof. The maximum achievable value of $N$ is $17$.

Let us denote the numbers on the two cards selected by Player $A$ in the $k$-th round as $a_k$ and $b_k$, where $a_k<b_k$. The number selected by Player $B$ is denoted as $c_k$, and the discarded number as $d_k$.

Let $T=d_1+d_2+d_3+d_4$. Then we have $S+T=1+2+\cdots +8=36$.

Player $B$ has a strategy to ensure $S-T\ge -3$, and consequently $S\ge 17$ (note that $S$ must be an integer).

If in the first round, $b_1-a_1\ge 4$, then Player $B$ selects $b_1$, and in the second round selects $a_2$. This gives $c_1-d_1\ge 4$ and $c_2-d_2\ge 1-8=-7$. If in the first round, $b_1-a_1\le 3$, then Player $B$ selects $a_1$, and in the second round selects $b_2$. This gives $c_1-d_1\ge -3$ and $c_2-d_2\ge 1$.

Therefore, Player $B$ can always ensure $(c_1+c_2)-(d_1+d_2)\ge -3$.

In the third and fourth rounds, Player $B$ can always ensure $(c_3+c_4)-(d_3+d_4)\ge 0$. This is because after Player $A$ has chosen $a_3$ and $b_3$, $a_4$ and $b_4$ are also determined. Player $B$ can then select the pair $(a_3,b_4)$ or $(b_3,a_4)$ that yields the larger sum. Thus, Player $B$ can always guarantee $S-T\ge -3$.

Player $A$ has a strategy to ensure $S\le 17$:

In the first round, Player $A$ selects $(3,6)$. If Player $B$ chooses $3$, then in the second round Player $A$ selects $(4,5)$. The sum of Player $B$'s selections in the first two rounds will not exceed $8$.

In the last two rounds, Player $A$ selects $(1,2)$ and $(7,8)$, and Player $B$'s selections in these rounds will not exceed $9$. Therefore, $S\le 17$.

If in the first round Player $B$ chooses $6$, then in the second round Player $A$ selects $(1,8)$, forcing Player $B$ to choose $1$. In the last two rounds, Player $A$ selects $(2,4)$ and $(5,7)$, and Player $B$'s selections in these rounds will not exceed $9$. Thus, $S\le 6+1+9=16$.

In conclusion, Player $A$ has a strategy to ensure $S\le 17$. $\square$